What is a constellation as astronomers define it today? What does it mean when an astronomer says, “I saw a comet in Orion last night”?

Answers

Answer 1

Answer:

Explanation:

Constellation: The complete sky has been divided in 88 different areas, in a way we have divided Earth in countries, not necessarily having same shapes and size. These 88 areas are known as constellations. These contains a lot of stars. When we join the brightest stars together we can imagine a shape out of them which is called as Asterism. Most of the people are unaware of this difference. Some of the famous constellations are Orion, Taurus, Gemini, Hydra, Ursa Major etc.

When an astronomer says that there is a comet is in the Orion, he means that a comet is in the boundaries of Orion constellation.

Answer 2

Answer:

Final answer:

A constellation is a group of stars that forms a certain pattern in the sky. 'Seeing a comet in Orion' means that the comet was observed in the region of the sky defined by the constellation Orion.

Explanation:

A constellation is a group of stars that astronomers have grouped together into patterns which represent certain symbols as a means of recognizing or remembering them. They are only human constructs not real associations of stars. When an astronomer says, “I saw a comet in Orion last night”, they are referring to a specific area in the sky defined by the constellation Orion. The constellation acted as a reference point that helped the astronomer locate and observe the comet.

Learn more about Constellation here:

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How much work would it take to push two protons very slowly from a separation of 2.00×10−10 m (a typical atomic distance) to 3.00×10−15 m (a typical nuclear distance)?

Answers

Electric potential is the amount of work needed to move a unit charge from a point to a specific point against an electric field. The work would it take to push two protons will be 7.7×10⁻¹⁴.

What is electric potential?

Electric potential is the amount of work needed to move a unit charge from a point to a specific point against an electricfield.

The given data in the problem is;

q is the charge= 1.6 ×10⁻¹⁹ C

V is the electric potential

r₁ is the first separation distance= 2.00×10−10 m

r₂ is the second separation distance= 3.00×10−15 m

The electric potential generated by the proton at rest at the two points, using the formula:

Firstly the electric potential at loction 1

$\rm V=(Kq)/(r) \n\n v_i= 9* 10^9 * (1.6*10^(-19))/(2.0*10^(-10))$

The electric potential at loction 2

$V_f = 9 * 10^9 (1.6 * 10^(-19))/(3.0*10^(-15)) \n\n \rm v_f= 4.8 *10^5 \ V$

The product of difference of electric potential and charge is defined as the workdone.

$\rm W= q \triangle V \n\n \rm W= 1.6 * 10^-19 *( 4.8*10^5 -7.2) \n\n \rm W= 7.7 * 10^(-14)$

Hence the work would it take to push two protons will be 7.7×10⁻¹⁴.

To learn more about electric potential work refer to the link.

brainly.com/question/12707371

We can visualize the problem in another way, which is equivalent but easier to solve: let's imagine we hold one proton in the same place, and we move the other proton from a distance of 2.00×10−10 m to a distance of 3.00×10−15 m from the first proton. How much work is done?

The work done is equal to the electric potential energy gained by the proton:

$W=q \Delta V$

where $q=1.6 \cdot 10^(-19)C$ is the charge of the proton and $\Delta V$ is the potential difference between the final position and the initial position of the proton. To calculate this $\Delta V$ , we must calculate the electric potential generated by the proton at rest at the two points, using the formula:

$V=k(Q)/(r)$

where $k=9.0 \cdot 10^9 N m^2 C^(-2)$ is the Coulomb constant and Q is the proton charge. Substituting the initial and final distance of the second proton, we find

$V_i = (9.0 \cdot 10^9 )(1.6 \cdot 10^(-19))/(2.0 \cdot 10^(-10))=7.2 V$

$V_f = (9.0 \cdot 10^9 )(1.6 \cdot 10^(-19))/(3.0 \cdot 10^(-15))=4.8 \cdot 10^5 V$

Therefore, the work done is

$W=q \Delta V=(1.6 \cdot 10^(-19)C)(4.8 \cdot 10^5 V-72 V)=7.7 \cdot 10^(-14) J$

wo parallel plates of area 100cm2are given charges of equal magnitudes 8.9 ×10−7C but opposite signs. The electric field within the dielectric material filling the space between the plates is 1.4 ×106V/m. (a) Calculate the dielectric constant of the material. (b) Determine the magnitude of the charge induced on each dielectric surface.

Answers

(a) 7.18

The electric field within a parallel plate capacitor with dielectric is given by:

$E=(\sigma)/(k \epsilon_0)$ (1)

where

$\sigma$ is the surface charge density

k is the dielectric constant

$\epsilon_0$ is the vacuum permittivity

The area of the plates in this capacitor is

$A=100 cm^2 = 100\cdot 10^(-4) m^2$

while the charge is

$Q=8.9\cdot 10^(-7)C$

So the surface charge density is

$\sigma = (Q)/(A)=(8.9\cdot 10^(-7) C)/(100\cdot 10^(-4) m^2)=8.9\cdot 10^(-5) C/m^2$

The electric field is

$E=1.4\cdot 10^6 V/m$

So we can re-arrange eq.(1) to find k:

$k=(\sigma)/(E \epsilon_0)=(8.9\cdot 10^(-5) C/m^2)/((1.4\cdot 10^6 V/m)(8.85\cdot 10^(-12) F/m))=7.18$

(b) $7.66\cdot 10^(-7)C$

The surface charge density induced on each dielectric surface is given by

$\sigma' = \sigma (1-(1)/(k))$

where

$\sigma=8.9\cdot 10^(-5) C/m^2$ is the initial charge density

k = 7.18 is the dielectric constant

Substituting,

$\sigma' = (8.9\cdot 10^(-5) C/m^2) (1-(1)/(7.18))=7.66\cdot 10^(5) C/m^2$

And by multiplying by the area, we find the charge induced on each surface:

$Q' = \sigma' A = (7.66\cdot 10^(-5) C/m^2)(100 \cdot 10^(-4)m^2)=7.66\cdot 10^(-7)C$

PLEASE HELP IT'S DUE IN LIKE 2 MINUTES

Answers

Answer:

1kg

Explanation:

this box is the smallest and weighs the least. Hope this helps :]

The speed of light in a transparent plastic is 2.5x108 m/s. If a ray of light in air(with n-1) strikes this plastic at an angle of incidence of 31.3 degrees, find the angle of the transmitted ray. Select one: a. 31.30 degrees b. 26.08 degrees c, 37.56 degrees d. 38.57 degrees e. 0.67 degrees

Answers

Answer:

Angle of transmitted ray is $29.14^(o)$

Explanation:

According to snell's law we have

$n_(1)sin(\theta _(i))=n_(2)sin(\theta r)$

Since the incident medium is air thus we have $n_(1)=1$

By definition of refractive index we have

$n=(c)/(v)$

c = speed of light in vacuum

v = speed of light in medium

Applying values we get

$n_(2)=(3* 10^(8))/(2.5* 10^(8))=1.2$

Thus using the calculated values in Snell's law we obtain

$sin(\theta _(r))=(sin(31.3)/(1.2)\n\n\therefore sin(\theta _(r))=0.4329\n\n\theta_(r)=sin^(-1)(0.4329)\n\n\theta _(r)=29.14$

Answer:

Angle made by the transmitted ray = 25.65°

Explanation:

Speed of light in plastic = v = 2.5 × 10⁸ m/s

refractive index of plastic (n₂) / refractive index of air (n₁)

= speed of light in air c / speed of light in plastic v.

⇒ n₂ = (3× 10⁸) / (2.5 × 10⁸) = 1.2

Angle of incidence = 31.3° = i

n₁ sin i = n₂ sin r

⇒ sin r = (1)(0.5195) / 1.2 = 0.4329

⇒ Angle made by the transmitted ray = r = sin⁻¹ (0.4329) =25.65°

Which volcanoes are formed by pyroclastic deposits? Select all that apply. A. cinder cone B. stratovolcano C. shield volcano

Answers

Answer:

Its a cinder cone cause after it all falls down to make deposits.

cinder cone and stratovolcano

A remote controlled car is moving in a vacant parking lot. The velocity of the car as a function of time is given by v= [5.00 m/s – (0.0180 m/s3)t^2 ]i+[2.00 m/s + (0.550 m/s2)t ]j .a) What are ax(t) and ay(t), the x- and y- components of cars acceleration as a function of time?
b) What are the magnitude and direction of the velocity of the car at t= 8 sec?
c) What is the magnitude and direction of cars acceleration at t=8 sec

Answers

Maybe try A for your answer

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